Question: A curve in the plane is defined parametrically by the equations $x=-2\ln(-t)$ and $y=2t^2+t$. Find the value of $\dfrac{dy}{dx}$ at $t=2$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{1}{9}$ (Choice B) B $-9$ (Choice C) C $-\dfrac{1}{10}$ (Choice D) D $10$
In general, to find the derivative (i.e. the expression for $\dfrac{dy}{dx}$ ) of a function defined parametrically by the equations $x=u(t)$ and $y=v(t)$ (where $u$ and $v$ are any functions of $t$ ), we use the following rule: $\dfrac{dy}{dx}=\dfrac{\left(\dfrac{dy}{dt}\right)}{\left(\dfrac{dx}{dt}\right)}=\dfrac{v'(t)}{u'(t)}$ We are given that $x=-2\ln(-t)$ and $y=2t^2+t$ : $\begin{aligned} \dfrac{dy}{dx}&=\dfrac{\dfrac{d}{dt}\left(2t^2+t\right)}{\dfrac{d}{dt}(-2\ln(-t))} \\\\ &=\dfrac{4t+1}{\left(-\dfrac{2}{t}\right)} \\\\ &=-\dfrac{4t^2+t}{2} \gray{\text{Simplify}} \end{aligned}$ Now let's evaluate $\dfrac{dy}{dx}$ at $t= {2}$ : $\begin{aligned} &\phantom{=}-\dfrac{4(2)^2+ 2}{2} \\\\ &=-9 \end{aligned}$ In conclusion, the value of $\dfrac{dy}{dx}$ at $t=2$ is $-9$.